physics

A student throws a ball vertically upward such that it travels 7.4m to its maximum height.

Part A
If the ball is caught at the initial height 2.7s after being thrown, what is the ball’s average speed?

Part B
What is the ball’s average velocity?

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  1. Part A
    d=r*t
    distance=rate*time

    7.4m=r*2.7s

    7.4m/2.7s=r

    r=2.7m/s

    Part B
    Use the following equation:

    Vf^2=Vi^2 +2 ad

    Where:

    Vf=0
    a=g=9.8m/s
    Vi=?
    d=7.4m

    0=Vi^2 +2(-9.8m/s^2)*(7.4m)

    0=Vi^2-145m^2/s^2

    145m^2/s^2=Vi^2

    sqrt*(145)=Vi

    Vi=12.0 m/s

    AverageV=(Vf+Vi)/2=(0+12.0m/s)/2=6m/s

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  2. (B) since the ball ended up where it started, its average velocity was 0

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  3. Steve you may be correct, but both questions can be worded a little bit better. For A: I wasn't sure if d=7.4m or d=2(7.4)=14.8m. If so, then r=14.8m/2.7s=5.5m/s. For B: I wasn't sure if they wanted to know the average velocity for the trip up, or the trip up and back down. If so, then the average velocity is 0.

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  4. Yeah. I've noticed that either textbooks have gotten very sloppy in their language, or students who post here aren't very careful in transcribing their assignments.

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  5. Yes, but the for A the speed is 5.5m/s and 0 for the velocity. I must have been tired. I just didn' t read the question carefully.

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